Problem 55 - Thomas Calculus in SI Units

Question

Find the limits $$\lim _{x \rightarrow 0^{+}} x^{-1 / \ln x}$$

Step-by-Step Solution

Answer

The limit is \(e^{-1}\).

Step 1: Rewrite the Expression

The given expression is \(x^{-1/\ln x}\). Start by rewriting this expression to make it easier to handle:\[ y = x^{-1/\ln x} = e^{\ln(x^{-1/\ln x})} = e^{-\frac{1}{\ln x} \ln x} = e^{-1} \] by properties of logarithms.

Step 2: Consider the Limit

Now, we need to find the limit of the constant expression as \(x\) approaches 0 from the positive side:\[ \lim_{x \to 0^+} x^{-1/\ln x} = \lim_{x \to 0^+} e^{-1} \]

Step 3: Evaluate the Limit

Since \(e^{-1}\) is a constant, the limit is simply \(e^{-1}\):\[ \lim_{x \to 0^+} x^{-1/\ln x} = e^{-1} \]

Step 4: Verify

Verify that the logic holds for approaching zero, as you initially did not need any substitution. The simplification used in Step 1 directly leads to the final constant value!

Key Concepts

Understanding Logarithmic Functions

Logarithmic functions are mathematical expressions involving logarithms. A logarithm answers the question of how many times we need to multiply a base to get a particular number. The most common logarithmic function is the natural logarithm, noted as \(\ln(x)\), which uses the base \(e\). The base \(e\) is an irrational number approximately equal to 2.71828. In our original exercise, understanding logarithms helps simplify the expression \(x^{-1/\ln x}\). Using properties of logarithms, like \(\ln(a^b) = b\ln(a)\), allows us to convert seemingly complex expressions into more manageable ones. These manipulations play a key role in calculus and finding limits, where handling potentially difficult expressions is often essential.

Basic Calculus Problems

Calculus problems often involve finding limits, derivatives, and integrals. Limits explore the behavior of a function as it approaches a specific point. The problem we tackled involved finding the limit of a transformed logarithmic function as \(x\) approached zero.

When solving calculus problems, especially with limits, several techniques can be applied:

• Rewriting expressions using identities or logarithmic properties.
• Employing substitution methods, though less in this specific problem due to a constant result.
• Utilizing L'Hôpital's Rule when faced with indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

This exercise shows how simplifying the expression to a constant—\(e^{-1}\)—bypasses more complex calculus methods, yet achieves the solution seamlessly.

Exploring Limits Approaching Zero

The concept of limits approaching zero develops our understanding of how functions behave near points of interest. Specifically, as \(x\) approaches 0 from the positive side (\(x \to 0^+\)), we observe what happens to the function's value.

In this particular exercise, we determined the behavior of \(x^{-1/\ln x}\) as \(x\) nears zero. The expression took advantage of constant values in limits, showing how manipulation in early steps set up for a straightforward resolution.

Here are some key aspects to consider when working with limits approaching zero:

• Direction matters: approaching from the left (negative) or right (positive) can yield different results.
• Expressions sometimes simplify to constants, bypassing the need for more elaborate calculation.
• Understanding that not all limits will resolve neatly is important—some may approach infinity or remain undefined.

These insights are crucial for comprehending the behavior of functions and how calculus can predict changes even as values shrink close to zero.